Integration by Observation

In integration by observation, we look at a product or a quotient to see a certain mathematical connection between its two parts.

Case 1

∫ f ^'(x) / f(x) dx = ln|f(x)| + c

If the numerator of the fraction is the differential coefficient of the denominator, then the integral is ln|denominator|

Proof:

∫ f '(x) / f(x) dx
let u = f(x) => du/dx = f '(x) => du = f '(x)dx
∫ f '(x) / f(x) dx = ∫ 1/u du
= ln|u| + c
= ln|f(x)| +c

E.g.1

∫ cot x dx
= ∫ cos x/sin x
= ln|sin x| + c

E.g.2

∫ tan x dx
= ∫ sin x/cos x
= - ∫ -sin x/cos x
= - ln|cos x| + c
or 0 - ln|cos x| = ln 1 - ln|cos x| = ln|1/cos x| = ln|sec x| + c

E.g.3

∫ x+1/x² + 2x + 3 dx
= 1/2 ∫ 2x+2/x² + 2x + 3 dx
= 1/2 ln|x² + 2x + 3| + c

E.g.4

∫ 1/tan x cos²x dx
= ∫ sec² x/tan x dx
= ln|tan x| + c

Case 2

∫ f '(x) [f(x)]ⁿ dx = [f(x)]ⁿ⁺¹/n+1 + c

Please note that f '(x) is the differential coefficient of just the inner function, not the whole function.

Proof:

∫ f '(x) [f(x)]ⁿ dx
let u = f(x) => du/dx = f '(x) => du = f '(x) dx
∫ f '(x) / [f(x)]ⁿ dx = ∫ uⁿ du
= uⁿ⁺¹/n+1 + c
= [f(x)]ⁿ⁺¹/n+1 + c

E.g.1

∫ sec² x tan⁶ x dx
f '(x) = sec² x and f(x) = tan x
∫ sec² x tan⁶ x dx
= tan⁷x / 7 + c

E.g.2

∫ (x + 1) (x² + 2x -3) dx
= 1/2 ∫ (2x + 2) (x² + 2x -3) dx
= (x² + 2x -3)²/2 + c

E.g.3

∫ (x - 1) √ (x² - 2x) dx
= 1/2 ∫ (2x - 2) √ (x² - 2x) dx
= 1/2 (x² - 2x)^3/2/3/2 + c = (x² - 2x)^3/2/3 + c

E.g.3

∫ cos³x dx
= ∫ cos x cos² x dx
= ∫ cos x (1 - sin²x) dx
= ∫ (cos x - cos x sin²x) dx
= sin x - sin³x / 3 + c